Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
An integer n is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the string representation of the integer n in base b is palindromic.
Given an integer n, return true if n is strictly palindromic and false otherwise.
A string is palindromic if it reads the same forward and backward.
Example 1:
Input: n = 9 Output: false Explanation: In base 2: 9 = 1001 (base 2), which is palindromic. In base 3: 9 = 100 (base 3), which is not palindromic. Therefore, 9 is not strictly palindromic so we return false. Note that in bases 4, 5, 6, and 7, n = 9 is also not palindromic.
Example 2:
Input: n = 4 Output: false Explanation: We only consider base 2: 4 = 100 (base 2), which is not palindromic. Therefore, we return false.
Constraints:
4 <= n <= 105Problem summary: An integer n is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the string representation of the integer n in base b is palindromic. Given an integer n, return true if n is strictly palindromic and false otherwise. A string is palindromic if it reads the same forward and backward.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Two Pointers
9
4
palindrome-number)stone-game)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2396: Strictly Palindromic Number
class Solution {
public boolean isStrictlyPalindromic(int n) {
return false;
}
}
// Accepted solution for LeetCode #2396: Strictly Palindromic Number
func isStrictlyPalindromic(n int) bool {
return false
}
# Accepted solution for LeetCode #2396: Strictly Palindromic Number
class Solution:
def isStrictlyPalindromic(self, n: int) -> bool:
return False
// Accepted solution for LeetCode #2396: Strictly Palindromic Number
impl Solution {
pub fn is_strictly_palindromic(n: i32) -> bool {
false
}
}
// Accepted solution for LeetCode #2396: Strictly Palindromic Number
function isStrictlyPalindromic(n: number): boolean {
return false;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.